Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m, i, j;
  cin >> n >> m;
  int a1[n][m], a2[n][m], b1[n], b2[m];
  vector<int> row1, col1, row2, col2;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> a1[i][j];
      a2[i][j] = a1[i][j];
    }
  }
  for (i = 0; i < n; i++) {
    b1[i] = a1[i][0];
    for (j = 1; j < m; j++) b1[i] = min(b1[i], a1[i][j]);
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < b1[i]; j++) row1.push_back(i + 1);
    for (j = 0; j < m; j++) a1[i][j] -= b1[i];
  }
  for (i = 0; i < m; i++) {
    for (j = 1; j < n; j++) {
      if (a1[0][i] != a1[j][i]) break;
    }
    if (j != n) break;
    for (j = 0; j < a1[0][i]; j++) col1.push_back(i + 1);
  }
  if (i != m)
    cout << -1;
  else {
    for (i = 0; i < m; i++) {
      b2[i] = a2[0][i];
      for (j = 1; j < n; j++) b2[i] = min(b2[i], a2[j][i]);
    }
    for (i = 0; i < m; i++) {
      for (j = 0; j < b2[i]; j++) col2.push_back(i + 1);
      for (j = 0; j < n; j++) a2[j][i] -= b2[i];
    }
    for (i = 0; i < n; i++) {
      for (j = 0; j < a2[i][0]; j++) row2.push_back(i + 1);
    }
    if (row1.size() + col1.size() <= row2.size() + col2.size()) {
      cout << row1.size() + col1.size() << "\n";
      for (auto i : row1) cout << "row " << i << "\n";
      for (auto i : col1) cout << "col " << i << "\n";
    } else {
      cout << row2.size() + col2.size() << "\n";
      for (auto i : col2) cout << "col " << i << "\n";
      for (auto i : row2) cout << "row " << i << "\n";
    }
  }
}